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AC 20161207 exercise 1
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子字串
請設計一個程式，使用者會輸入兩個字串，兩個字串間用一個空格隔開，每個字串的內容只會包含大小寫的英文字母，請判斷第二個字串是否出現在第一個字串內。請將大小寫視為同一個字。
輸入範例1：
friEND end
輸出範例1：
There is "end" in "friEND"
輸入範例2：
friEND friends
輸出範例2：
There are no "friends" in "friEND"
asked
Dec 7, 2016
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2016-1 程式設計(一)AC
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Shun-Po
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answered
Dec 18, 2016
by
410523016
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answered
Dec 18, 2016
by
410523016
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Hidden content!#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdlib.h>
int main()
{
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answered
Dec 18, 2016
by
Gulang
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-122
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Hidden content!#include <stdio.h>
#include ** * * * **
#include <ctype.h>
int main()
{
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answered
Dec 18, 2016
by
Gulang
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-122
points)
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Hidden content!#include<stdio.h>
#include<string.h>
#include<ctype.h>
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{
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** * ** ** * * * * ** * * 0;
}
answered
Dec 18, 2016
by
Jim870729
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-102
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Hidden content!#include<stdio.h>
#include<string.h>
#include<ctype.h>
int main ( void )
{
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}
answered
Dec 18, 2016
by
Jim870729
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Hidden content!#include<stdio.h>
* * ** ***
#include<ctype.h>
int main ( void )
{
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* ** * * ** ** * ** 0;
}
answered
Dec 18, 2016
by
Jim870729
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-102
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Hidden content!#include<stdio.h>
#include<string.h>
#include<ctype.h>
int main ( void )
{
* * * ** * * ** i,j;
** * ** * * ** ** ** *** * *** * ** * ** * * * * * * * * *
***** * * * * * * * * * * * * * * *
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* * ** ** ** ** * *
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* * * * * ** * **** *** ** * * * *** *** ****** ** *** ** ** = c2;
***** * * * ** **** * * * ***
**** *** * * ** * *** * * = strstr(a,b);
* *** ** * ** * * * * *
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* * * ** * * * * * ** * ** *** *** ** * * * * ** * ** ** * is %s in %s",b,a);
** * * ** *** * * * * * *** 0;
}
answered
Dec 18, 2016
by
Jim870729
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-102
points)
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Hidden content!#include<stdio.h>
#include<string.h>
#include<ctype.h>
int main ( void )
{
* * * ** * ** * *** ** * i,j;
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* ** * **** * * * ** * * * * * ** ** * * * = c2;
** * * ***** * * * * ** * *
** * * *** * * * * * ** *** * * = strstr(a,b);
* * * * * * * * * *** * **
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* * * ** * * * ** * * ** **
* ** * * * * *** ** * * * * * ** ** ** * * * * * * * * * is %s in %s",b,a);
** * * ** ** * * ** * * ** * 0;
}
answered
Dec 18, 2016
by
Jim870729
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-102
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Hidden content!#include ** ** * * *
#include * ** * **
#include * * ** * * *
#include *** * * * * * *
#include ** ** * * ****
int main(void){
* **** * * ** ** ** * ****** * * * * * *
* *
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* ** * ** * ** * ** * * ** * *
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** * *
* ** ** ** * *
* * ** * * * ** * * * * * * ++;
* * ** * * ** * * * * ** *** *
}
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**** *
* * * * * * ** * * *** * * * *
**** * * *** * ** * ++;
* *** * * * *
}
* * * * *** ** * ** * * * * ** * ** * ***
** ** ** * * * * * * * *
* * * * (i = 0;i <= sizes;i++)
***
* * * * *
* * ** * = s[i];
** **
* * * *
* * * (i = 0;i <= sizet;i++)
* * * **
** * * * * **
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**** * *
** * *
* ** * ** * * * ** ** * * * * *
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{
* ** * ***** * **
** * * ** ** * *
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* ** * * * **
* ** * **
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}
* **** **** ** * ***** * ** * ** **** *
for (i = 0;i <= sizet;i++)
{
* * *
* * **** * * * *
* *** ** *** * ** = f[i] - 32;
** * *
** * * * *
* ** * * * *** * * = f[i];
* * * * * *
}
* * * * * * * *** **** * ** **
//asplpSLP slp
** * * (i = 0;p2[i] != '\0';i++)
* * * *
* * ** * (p2[i] == f2[0])
* ** * * * **
** * * * * *
* ** * *** ** ** **** * **** = i;
** ** ** ** * * * ***
* * ** * * *** ** * (u = 0;u < sizet;u++)
* ** *** * ** *
*** *** * ** ** * ** ** *
* ** * * * * * * * * * * * **
* * ** * * * ** * (p2[j+u] == f2[u])
* ** ** ** * ** **** ** * ****** * * ***
* * ***** * ** * *** * * += 1;
* * * * ** * * ** * ** ** ** *
*** * ** **** * *** * ** ** *
** * * * * *** * * * * * * *
** * * * *** * ****** * * * *
** * ** ** * ** * * ** *** * * **
** * * **** * * * * * * ** = 0;
* * * * *** * ** *** * *** **** *
** * ** **** ** ** ** * * *
* * * ** ** ** * * * * ** * *
*** ** * * * * * *** ** *** * * *
* * ** * *** ** ** ** ** *** *** *
* ** * * * * * ** * *
* * * * * * *
*** * * * ** ** *
* ****** * **
* * *
** * * * * * * **** * * (d > 0)
* ** ** ** * * * * *** * *
** *** * * * * * * * * * * * * * ** is * * * * in * * * ** ** * * ***** ***
** * * ** * * * * * * * **** * * * * * **
* * *** * * * * * *
* * ** * ** * * * * * *
*** * * * * ** ** ** * * * are no * * * *** * * ** in ** * * * *** * * ** **** *
* * * ** ** *** * ** * * * 0;
**** * *** ** *** ****
}
answered
Dec 10, 2016
by
蘇瑋朝
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